Cauchy-Schwarz inequality

Non-negativity. For all $t\in\mathbb{R}$,
$$\|u-tv\|^2=\|u\|^2-2t\langle u,v\rangle+t^2\|v\|^2\ge0.$$

Case $v=0$. Then $\langle u,v\rangle=0$ and the inequality is trivial.

Case $v\neq0$. Choose $t=\langle u,v\rangle/\|v\|^2$:
$$\|u\|^2 - \frac{\langle u,v\rangle^2}{\|v\|^2}\ge0,$$
i.e. $\langle u,v\rangle^2\le\|u\|^2\|v\|^2$.

Conclusion. $|\langle u,v\rangle|\le\|u\|\|v\|$, with equality iff $u,v$ are linearly dependent.